3 ms·

I stopped reading at this point: > Roughly half of your measurements will be above average, and the other half below it. This is simply not true for most defi

I stopped reading at this point: > Roughly half of your measurements will be above average, and the other half below it. This is simply not true for most definitions of "average" (i.e., for all definitions of "average" except the median). Example: > (data <- c(1:10,100)) [1] 1 2 3 4 5 6 7 8 9 10 100 > sum(as.numeric(data > mean(data)))/length(data) [1] 0.09090909 A 90/10 split is hardly "roughly half of your measurements".

kqr 3y ago
Would it help you get past that point if it had said "of a stable process"?
- fjkdlsjflkds 3y ago
  Well... what do you mean by "a stable process" in this context? Let's try repeating the same example, but now drawing samples from a fixed distribution (in this case, a log-normal distribution): > data <- exp(rnorm(100)) > sum(as.numeric(data > mean(data)))/length(data) [1] 0.32 So, again, quite far from a 50/50 split, even though I am assuming a stable/fixed data generation process. In general, it would help if statistical subjects are not presented in a careless way (i.e., containing things which are obviously not true). I would suggest at least adding an "assuming a symmetrical distribution" (so that at least your claim is approximately correct under the arithmetic average and for bounded variance distributions). EDIT: If by "a stable process" you mean "a process following a stable distribution"... then, no, it doesn't help. Here's an example with samples drawn from a Lévy distribution (which is a stable distribution): > data <- rmutil::rlevy(100) > sum(as.numeric(data > mean(data)))/length(data) [1] 0.07
  - kqr 3y ago
    In this context "stable" means the thing it means in statistical process control, i.e. the operational definition of no measurements outside of 2.66 times the mean consecutive difference between observations. It is a problem -- particularly for software -- that SPC tools do not work with subexponential distributions, but it's separate from the observation that when SPC determines that a process is stable, rougly half of measurements will lie above the average.
    - shadowsun7 3y ago
      To be fair to OP, Wheeler never claims that for stable/in-control/predictable processes roughly half of the measurements will lie above the average. The only claim he makes is that 97% of all data points for a stable process (assuming the process draws from a J-curve or single-mound distribution) will fall between the limit lines. He can't make this claim (about ~half falling above/below the average line), because one of the core arguments he makes is that XmR charts are usable even when you're not dealing with normal distributions. He argues that the intuition behind how they work is that they detect the presence of more than one probability distribution in the variation of a time series. Some links below: Arguments for non-normality: https://spcpress.com/pdf/DJW220.pdf https://spcpress.com/pdf/DJW220.pdf https://www.spcpress.com/pdf/DJW354.Sep.19.The%20Normality-Myth.pdf https://www.spcpress.com/pdf/DJW354.Sep.19.The%20Normality-M... Claim of homogeneity detection: https://www.spcpress.com/pdf/DJW204.pdf https://www.spcpress.com/pdf/DJW204.pdf
      - kqr 3y ago
        I don't have the stats-fu to back it up but I would be very surprised if someone could point to a process where XmR charts are useful, but where the mean is not within 10–20 percentiles of the median.
    - fjkdlsjflkds 3y ago
      > the operational definition of no measurements outside of 2.66 times the mean consecutive difference between observations Not even a simple Gaussian distribution can hold up to this standard of "stability" (unless I understood incorrectly what you mean here): > data <- rnorm(1000) # i.i.d. normal data > mcd <- 2.66*mean(abs(diff(data))) # mean consecutive difference * 2.66 > sum(as.numeric(abs(data) > mcd))/length(data) # fraction of bad points [1] 0.002 Unless you are willing to add additional conditions (e.g., symmetry), I still don't see how criteria that pertain to variance and kurtosis (e.g., "the operational definition of no measurements outside of 2.66 times the mean consecutive difference between observations") can imply any strong relationship between the (sample) arithmetic mean (or any other mean) and the (population) median. In fact, even distributions for which the "arithmetic mean is approximately equal to the median" claim is roughly correct will almost certainly not display the same property when you use some other mean (e.g., geometric or harmonic mean). Either way, if you have some reference that supports the stated claim, I will be very happy to take a look at it (and educate myself in the process).
  - qznc 3y ago
    Draw an control chat with your data. Here is one for your initial example: http://beza1e1.tuxen.de/spc?data=1,2,3,4,5,6,7,8,9,10,100 http://beza1e1.tuxen.de/spc?data=1,2,3,4,5,6,7,8,9,10,100 It is immediately obvious that it is not a stable process.
    - kqr 3y ago
      I think GPs point is that there are software processes that are fundamentally stable but still generate values like that. I'm in the process of writing an article on that topic, because it annoys me I don't have a good answer.
    - fjkdlsjflkds 3y ago
      Ok. So, let's try again... First, let's draw some data and show that the sample mean is not roughly close to the sample median: > data <- asinh(exp(rnorm(30))) > mean(data) [1] 1.227973 > median(data) [1] 1.059046 > sum(as.numeric(data > mean(data)))/length(data) [1] 0.4 Now you can take the dataset and make a control chart with it: > cat(paste0(data,collapse=",")) 0.397291781860484,1.01791591678607,2.29127398581896,0.317548192016798,0.825779972770721,0.978034869623426,1.45689922574378,0.722379000865545,2.68231132467641,0.786713029297768,1.20492120955161,1.7082762081373,0.632259821911453,0.346590855307735,2.48238023470879,0.0989934260605276,1.61233320755675,0.906918026775941,1.73743152912329,1.21715325934946,2.78776306537914,0.296838101056961,2.29303061152949,2.65277514999252,0.88486942904647,0.0860402641329708,1.255123685342,0.526097043743278,1.53307173756969,1.10017654868633 https://beza1e1.tuxen.de/spc?data=0.397291781860484,1.01791591678607,2.29127398581896,0.317548192016798,0.825779972770721,0.978034869623426,1.45689922574378,0.722379000865545,2.68231132467641,0.786713029297768,1.20492120955161,1.7082762081373,0.632259821911453,0.346590855307735,2.48238023470879,0.0989934260605276,1.61233320755675,0.906918026775941,1.73743152912329,1.21715325934946,2.78776306537914,0.296838101056961,2.29303061152949,2.65277514999252,0.88486942904647,0.0860402641329708,1.255123685342,0.526097043743278,1.53307173756969,1.10017654868633 https://beza1e1.tuxen.de/spc?data=0.397291781860484,1.017915... Looks like the data is "stable" (according to that website) and, yet, you get a 40/60 split, rather than a 50/50 split.
      - qznc 3y ago

kqr 3y ago

Would it help you get past that point if it had said "of a stable process"?

fjkdlsjflkds 3y ago
Well... what do you mean by "a stable process" in this context? Let's try repeating the same example, but now drawing samples from a fixed distribution (in this case, a log-normal distribution): > data <- exp(rnorm(100)) > sum(as.numeric(data > mean(data)))/length(data) [1] 0.32 So, again, quite far from a 50/50 split, even though I am assuming a stable/fixed data generation process. In general, it would help if statistical subjects are not presented in a careless way (i.e., containing things which are obviously not true). I would suggest at least adding an "assuming a symmetrical distribution" (so that at least your claim is approximately correct under the arithmetic average and for bounded variance distributions). EDIT: If by "a stable process" you mean "a process following a stable distribution"... then, no, it doesn't help. Here's an example with samples drawn from a Lévy distribution (which is a stable distribution): > data <- rmutil::rlevy(100) > sum(as.numeric(data > mean(data)))/length(data) [1] 0.07
- kqr 3y ago
  In this context "stable" means the thing it means in statistical process control, i.e. the operational definition of no measurements outside of 2.66 times the mean consecutive difference between observations. It is a problem -- particularly for software -- that SPC tools do not work with subexponential distributions, but it's separate from the observation that when SPC determines that a process is stable, rougly half of measurements will lie above the average.
  - shadowsun7 3y ago
    To be fair to OP, Wheeler never claims that for stable/in-control/predictable processes roughly half of the measurements will lie above the average. The only claim he makes is that 97% of all data points for a stable process (assuming the process draws from a J-curve or single-mound distribution) will fall between the limit lines. He can't make this claim (about ~half falling above/below the average line), because one of the core arguments he makes is that XmR charts are usable even when you're not dealing with normal distributions. He argues that the intuition behind how they work is that they detect the presence of more than one probability distribution in the variation of a time series. Some links below: Arguments for non-normality: https://spcpress.com/pdf/DJW220.pdf https://spcpress.com/pdf/DJW220.pdf https://www.spcpress.com/pdf/DJW354.Sep.19.The%20Normality-Myth.pdf https://www.spcpress.com/pdf/DJW354.Sep.19.The%20Normality-M... Claim of homogeneity detection: https://www.spcpress.com/pdf/DJW204.pdf https://www.spcpress.com/pdf/DJW204.pdf
    - kqr 3y ago
      I don't have the stats-fu to back it up but I would be very surprised if someone could point to a process where XmR charts are useful, but where the mean is not within 10–20 percentiles of the median.
  - fjkdlsjflkds 3y ago
    > the operational definition of no measurements outside of 2.66 times the mean consecutive difference between observations Not even a simple Gaussian distribution can hold up to this standard of "stability" (unless I understood incorrectly what you mean here): > data <- rnorm(1000) # i.i.d. normal data > mcd <- 2.66*mean(abs(diff(data))) # mean consecutive difference * 2.66 > sum(as.numeric(abs(data) > mcd))/length(data) # fraction of bad points [1] 0.002 Unless you are willing to add additional conditions (e.g., symmetry), I still don't see how criteria that pertain to variance and kurtosis (e.g., "the operational definition of no measurements outside of 2.66 times the mean consecutive difference between observations") can imply any strong relationship between the (sample) arithmetic mean (or any other mean) and the (population) median. In fact, even distributions for which the "arithmetic mean is approximately equal to the median" claim is roughly correct will almost certainly not display the same property when you use some other mean (e.g., geometric or harmonic mean). Either way, if you have some reference that supports the stated claim, I will be very happy to take a look at it (and educate myself in the process).
- qznc 3y ago
  Draw an control chat with your data. Here is one for your initial example: http://beza1e1.tuxen.de/spc?data=1,2,3,4,5,6,7,8,9,10,100 http://beza1e1.tuxen.de/spc?data=1,2,3,4,5,6,7,8,9,10,100 It is immediately obvious that it is not a stable process.
  - kqr 3y ago
    I think GPs point is that there are software processes that are fundamentally stable but still generate values like that. I'm in the process of writing an article on that topic, because it annoys me I don't have a good answer.
  - fjkdlsjflkds 3y ago
    Ok. So, let's try again... First, let's draw some data and show that the sample mean is not roughly close to the sample median: > data <- asinh(exp(rnorm(30))) > mean(data) [1] 1.227973 > median(data) [1] 1.059046 > sum(as.numeric(data > mean(data)))/length(data) [1] 0.4 Now you can take the dataset and make a control chart with it: > cat(paste0(data,collapse=",")) 0.397291781860484,1.01791591678607,2.29127398581896,0.317548192016798,0.825779972770721,0.978034869623426,1.45689922574378,0.722379000865545,2.68231132467641,0.786713029297768,1.20492120955161,1.7082762081373,0.632259821911453,0.346590855307735,2.48238023470879,0.0989934260605276,1.61233320755675,0.906918026775941,1.73743152912329,1.21715325934946,2.78776306537914,0.296838101056961,2.29303061152949,2.65277514999252,0.88486942904647,0.0860402641329708,1.255123685342,0.526097043743278,1.53307173756969,1.10017654868633 https://beza1e1.tuxen.de/spc?data=0.397291781860484,1.01791591678607,2.29127398581896,0.317548192016798,0.825779972770721,0.978034869623426,1.45689922574378,0.722379000865545,2.68231132467641,0.786713029297768,1.20492120955161,1.7082762081373,0.632259821911453,0.346590855307735,2.48238023470879,0.0989934260605276,1.61233320755675,0.906918026775941,1.73743152912329,1.21715325934946,2.78776306537914,0.296838101056961,2.29303061152949,2.65277514999252,0.88486942904647,0.0860402641329708,1.255123685342,0.526097043743278,1.53307173756969,1.10017654868633 https://beza1e1.tuxen.de/spc?data=0.397291781860484,1.017915... Looks like the data is "stable" (according to that website) and, yet, you get a 40/60 split, rather than a 50/50 split.
    - qznc 3y ago